At 25°C one mole of acetic acid was allowed to react with one mole of ethyl alcohol until equilibrium was established. The equilibrium mixture was found to contain 0.2 mole of unused acid. Calculate the equilibrium constant of the reaction at the same temperature.
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The given reaction is :-
C
2
H
5
OH+CH
3
COOH⇌CH
3
COOC
2
H
5
+H
2
O
1 1 0 0
At eq. : (1−x) (1−x) x x
Given that, at equilibrium-
Moles of CH
3
COOC
2
H
5
=0.666
⇒x=0.666
Now, number of moles of C
2
H
5
OH=0.334
number of moles of CH
3
COOH=1−0.666=0.334
∴K
C
=
[C
2
H
5
OH][CH
3
COOH]
[CH
3
COOC
2
H
5
][H
2
O]
=
0.334×0.334
0.666×0.666
≈2×2≈4
please make as Brilliant
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